107年高考三級-工程數學詳解

107年公務人員高等考試三級考試試題

類 科 ：電力工程、電子工程、電信工程
科 目：工程數學

last modified: 8/21/2019

解： $$假設A與B相以，則A=P^{-1}BP$$ (一) $$P^{-1}(\lambda I+B)P= P^{-1}\lambda IP+P^{-1}BP = \lambda P^{-1}IP+A=\lambda I+A \Rightarrow \lambda I+A與\lambda I+B相似$$ (二) $$A=P^{-1}BP \Rightarrow tr(A)=tr(P^{-1}BP)=tr((P^{-1}B)(P))=tr((P)(P^{-1}B))=tr(B)$$ (三) $$A=P^{-1}BP \Rightarrow det(A)=det(P^{-1}BP) =det(P^{-1})det(B)det(P)= det(P^{-1})det(P)det(B) \\  =det(B)；   因此A,B相以，則  det(A)=det(B)；\\由(一)知:   \lambda   I+A與\lambda   I+B相似，所以det(\lambda   I+A)=det(\lambda   I+B)$$

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解： $$y=\sum_{m=0}^\infty {a_mx^m} =a_0+a_1x +a_2x^2+a_3x^3+\cdots+a_nx^n+\cdots\\ \Rightarrow y'=a_1+2a_2x+3a_3x^2 +\cdots+na_nx^{n-1}+\cdots\\ \Rightarrow xy'=a_1x+2a_2x^2+3a_3x^3 +\cdots+na_nx^{n}+\cdots\\ \Rightarrow (x+1)y'=a_1+(a_1+2a_2)x+(2a_2+3a_3)x^2+ \cdots+(na_n+(n+1)a_{n+1} )x^{n}+\cdots\\ (x+1)y'=y \Rightarrow \begin{cases} a_0=a_1\\ a_1=a_1+2a_2 \\ a_2=2a_2+3a_3\\ \cdots \\a_n=na_n+(n+1)a_{n+1} \end{cases} \Rightarrow \begin{cases} a_0=a_1\\ a_n=0,n\ge 2 \end{cases} \Rightarrow \bbox[red,2pt]{y=a_0+a_0x},a_0為常數$$

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解： $$f(x)在區間-\pi\le x\le \pi 的傅立葉級數為f(x)=a_0 +\sum_{n=1}^\infty{a_n\cos{n x}} + \sum_{n=1}^\infty{b_n\sin{n x}}\\ 其中 \begin{cases} a_0={1\over 2\pi}\int_{-\pi}^\pi{f(x)dx} = {1\over 2\pi}\int_{-\pi}^\pi{ {x^2\over 2}dx} = {1\over 4\pi} \left. \left[ {1\over 3}x^3 \right]\right|_{-\pi}^\pi ={1\over 6}\pi^2 \\ a_n={1\over \pi}\int_{-\pi}^\pi{f(x)\cos{nx}dx} = {1\over \pi}\int_{-\pi}^\pi{{x^2\over 2}\cos{nx}dx}\\  \;\;\;\; = {1\over 2\pi} \left . \left[ {x^2\over n}\sin{nx}+{2x\over n^2}\cos{nx} -{2\over n^3}\sin{nx} \right] \right|_{-\pi}^\pi = {2\over n^2}(-1)^n \\ b_n={1\over \pi}\int_{-\pi}^\pi{f(x)\sin{nx}dx} = {1\over \pi}\int_{-\pi}^\pi{{x^2\over 2}\sin{nx}dx} = 0(奇函數) \end{cases} \\ \Rightarrow f(x)={\pi^2 \over 6} + \sum_{n=1}^\infty{{2\over n^2}(-1)^n\cos{nx}} \Rightarrow f(\pi)={\pi^2 \over 6} + \sum_{n=1}^\infty{{2\over n^2}(-1)^n\cos{n\pi}} ={\pi^2 \over 6} + \sum_{n=1}^\infty{{2\over n^2}} \\= {\pi^2 \over 6} + 2({1\over 1^2}+ {1\over 2^2}+ {1\over 3^2}+ \cdots)\Rightarrow  f(\pi)={\pi^2\over 2}={\pi^2 \over 6} + 2({1\over 1^2}+ {1\over 2^2}+ {1\over 3^2}+ \cdots) \\\Rightarrow {1\over 1^2}+ {1\over 2^2}+ {1\over 3^2}+ \cdots= ({\pi^2\over 2}-{\pi^2 \over 6})\div 2= {\pi^2\over 6}\\ \Rightarrow 1+{1\over 4}+{1\over 9}+{1\over 16}+\cdots = {\pi^2\over 6}$$

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解： $$\int_C{f(z)dz} =\int_C{{3z^3+2 \over (z-1)(z^2+9)}dz} \Rightarrow \begin{cases}\text{Res}(f,z=1)=\left. {3z^3+2\over z^2+9} \right|_{z=1} = {5\over 10}={1\over 2} \\ \text{Res}(f,z=3i)=\left. {3z^3+2\over (z-1)(z+3i)} \right|_{z=3i} = {2-81i \over -18-6i} \\\text{Res}(f,z=-3i)=\left. {3z^3+2\over (z-1)(z-3i)} \right|_{z=-3i} = {2+81i \over -18+6i} \end{cases} \\ \Rightarrow \int_C{f(z)dz} =2\pi i\times\left(\text{Res}(f,z=1) +\text{Res}(f,z=3i)+\text{Res}(f,z=-3i) \right)\\=2\pi i\times \left({1\over 2} +{2-18i \over -18-6i}+ {2+18i \over -18+6i}\right)= 2\pi i \times \left({1\over 2}+{900\over 360} \right)=2\pi i\times 3= \bbox[red, 2pt]{6\pi i}$$

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乙、測驗題部分：(50分)

解： $$\begin{cases} A=\begin{bmatrix} 1&2\\ 3& 4 \end{bmatrix}\\ B=\begin{bmatrix} -1&-2\\ -3& -4 \end{bmatrix} \end{cases}  \Rightarrow \begin{cases} |A+B|=\begin{vmatrix} 0&0\\ 0& 0 \end{vmatrix}=0\\ |A|+|B|=\begin{vmatrix} 1&2\\ 3& 4 \end{vmatrix} + \begin{vmatrix} -1& -2\\ -3& -4 \end{vmatrix}=-2-2=-4\end{cases} \\ \Rightarrow |A+B|\ne |A|+|B|，故選\bbox[red,2pt]{(B)}$$

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解： $$\begin{cases} a=(1,1)\ne 0\\ b=(2,2)\ne 0 \end{cases} \Rightarrow a\times b=0，故選\bbox[red,2pt]{(D)}$$

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解： $$\left[\begin{matrix}1&a& -1 \\0&1&1 \\2 & 0 & 2 \end{matrix}\right] \equiv \left[\begin{matrix} \vec{u}\\ \vec{v}\\ \vec{w} \end{matrix}\right] \Rightarrow \vec{u}=m\vec{v}+n\vec{w},m,n為常數\Rightarrow \begin{cases} 2n=1\\ m=a \\ m+2n=-1 \end{cases} \\ \Rightarrow \begin{cases} n=1/2\\ m=-2 \end{cases} \Rightarrow a=m=-2，故選\bbox[red,2pt]{(A)}$$

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解： $$rank(A)=n \Rightarrow nullity(A)=n-n=0，故選\bbox[red,2pt]{(D)}$$

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解： $$L(x,y,z)=(x-2y, 2x+y) \equiv \left[\begin{matrix}1&-2&0 \\2&1&0 \end{matrix}\right]\left[\begin{matrix}x\\y\\z \end{matrix}\right]，故選\bbox[red,2pt]{(D)}$$

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解： $$\left[\begin{matrix} -1 & 2 & 0 & 4 & 5 &-3 \\ 3 & -7 & 2 & 0 & 1 & 4\\ 2 & -5 & 2 & 4 & 6 &1 \\4 & -9 & 2 & -4 & -4 & 7 \end{matrix} \right] \xrightarrow{3r_1+r_2,\;2r_1+r_3\;,4r_1+r_4} \left[\begin{matrix} -1 & 2 & 0 & 4 & 5 &-3 \\ 0 & -1 & 2 & 12 & 16 & -5\\ 0 & -1 & 2 & 12 & 16 &-5 \\0 & -1 & 2 & 12 & 16 & -5 \end{matrix} \right]\\ \xrightarrow{-r_2+r_3,\,-r2+r_4}\left[\begin{matrix} -1 & 2 & 0 & 4 & 5 &-3 \\ 0 & -1 & 2 & 12 & 16 & -5\\ 0 & 0 & 0 & 0 & 0 &0 \\0 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right]\Rightarrow \text{Rank}=2，故選\bbox[red,2pt]{(B)}$$

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解： $$e^{5+2i}= e^5\cdot e^{2i} =e^5\cdot (\cos{2}+i\sin{2})，故選\bbox[red,2pt]{(C)}$$

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解： $$f(z)={\sinh{z}\over z^2} \Rightarrow \text{Res}(f,z=0) =\left. {d\over dz}\sinh{z} \right|_{z=0} =\left. {d\over dz}{e^z-e^{-z}\over 2} \right|_{z=0} = \left. {e^z+e^{-z}\over 2} \right|_{z=0} ={1+1\over 2}=1\\ \Rightarrow \int_C{\sinh{z}\over z^2}dz =2\pi i\times \text{Res}(f,z=0)= 2\pi i\times 1=2\pi i，故選\bbox[red,2pt]{(A)}$$

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解： $$由公式:L^{-1}\{F(s)\} =f(t) \Rightarrow L^{-1}\{e^{-as}F(s)\} =u(t-a)f(t-a)\\ 因此 L^{-1}\left\{{e^{-2s}\over s^2-3s+2} \right\}=  L^{-1}\left\{e^{-2s}\left( {1\over s-2}-{1\over s-1}\right)\right\} =u(t-2)\left(e^{2(t-2)}- e^{1(t-2)}\right) \\ =\left( -e^{t-2}+ e^{2t-4} \right) u(t-2)，故選\bbox[red,2pt]{(B)}$$

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解： $$y(t)-\int_0^t{y(\tau)(t-\tau)d\tau}=2-{1\over 2}t^2 \Rightarrow y(0)-\int_0^0{y(\tau)(t-\tau)d\tau}=2-{1\over 2}0^2 \Rightarrow y(0)=2\\ 又y(t)=a+be^t+ce^{-t} \Rightarrow y(0)=a+b+c =2，故選\bbox[red,2pt]{(C)}$$

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解： $$2+\left(6x-e^{-2y} \right){dy\over dx}=0 \Rightarrow 2dx+\left(6x-e^{-2y} \right)dy=0 \equiv Mdx+Ndy=0 \\ \Rightarrow \begin{cases} M=2\\ N=6x-e^{-2y} \end{cases} \Rightarrow f(y)= {{\partial M\over \partial y} -{\partial N \over \partial x}\over M} = {0-6\over2} = -3 \\\Rightarrow 積分因子I=e^{-\int{f(y)dy}} =e^{\int{3dy}} =e^{3y}，故選\bbox[red,2pt]{(C)}$$

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解： $$\begin{cases} x'(t)=-2y(t) \\ y'(t)={1\over 2}x(t)\end{cases} \Rightarrow \begin{cases} x''(t)=-2y'(t)=-2({1\over 2}x(t)) \\ y''(t)={1\over 2}x'(t)={1\over 2}(-2y(t))\end{cases} \Rightarrow \begin{cases} x''(t)+x(t)=0 \\ y''(t)+y(t) =0 \end{cases} \\ \Rightarrow \begin{cases} x(t)=A\cos{t}+B\sin{t} \\ y(t)=C\cos{t}+D\sin{t}\end{cases} 代回\begin{cases} x'(t)=-2y(t) \\ y'(t)={1\over 2}x(t)\end{cases}可得\begin{cases} B+2C=0 \\ A-2D=0\end{cases}\\ 由\begin{cases} x(0)=2 \\ y(0)=0\end{cases} \Rightarrow \begin{cases} A=2 \\ C=0\end{cases}\Rightarrow \begin{cases} D=1 \\ B=0\end{cases}\Rightarrow \begin{cases} x(t)=2\cos{t} \\ y(t)=\sin{t}\end{cases}，故選\bbox[red,2pt]{(A)}$$

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解： $$(x-1)^2y''-4xy'+4y'+4y=0 \Rightarrow (x-1)^2y''-4(x-1)y'+4y=0\\ 令u=x-1,則上式為u^2y''-4uy'+4y=0\\ 令y=u^m \Rightarrow y'=mu^{m-1} \Rightarrow y''=m(m-1)u^{m-2} \Rightarrow m(m-1)u^m-4mu^m+4u^m=0\\ \Rightarrow m(m-1)-4m+4=0 \Rightarrow m^2-5m+4=0 \Rightarrow (m-4)(m-1)=0 \Rightarrow m=4,1\\ \Rightarrow y=c_1u^4+c_2u^1 = c_1(x-1)^4+c_2(x-1)，故選\bbox[red,2pt]{(B)}$$

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解： $$依定義，故選\bbox[red,2pt]{(B)}$$

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解： $$令u_x=P,則u_{xy}=4u_x \Rightarrow {d P\over d y}=4P \Rightarrow {dp\over P}=4dy \Rightarrow \ln{P}=4y+A(x) \Rightarrow P= B(x)e^{4y} \\ \Rightarrow u=\int{P dx} =\int {B(x)e^{4y}\,dx}= e^{4y}\int{B(x)\,dx}=e^{4y}C(x)+D(y)，故選\bbox[red,2pt]{(C)}$$

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解： $$\begin{cases} L\{\sinh{at}\}={a\over s^2-a^2} \\ L\{\cosh{at}\} ={s\over s^2-a^2} \end{cases} \Rightarrow L^{-1}\{F(s)\} = L^{-1}\{{5s+1\over s^2-25}\} = L^{-1}\{{5s\over s^2-5^2} +{1\over s^2-5^2}\} \\ =5L^{-1}\{{s\over s^2-5^2}\}+ {1\over 5} L^{-1}\{ {5\over s^2-5^2}\} = 5\cosh{5t}+ {1\over 5}\sinh{5t}，故選\bbox[red,2pt]{(B)}$$

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解： $$選項(A)正確 \Rightarrow 選項(B)錯誤，故選\bbox[red,2pt]{(B)}$$

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解： $$不合乎條件的事件A=\{(x,y) \mid x\ne y 且 x\times y 是奇數\}= \{ (1,3), (1,5), (3,1) ,(3,5), (5,1), (5,3)\}\\ \Rightarrow \#(A)=6 \Rightarrow 1-P(A)=1-6/36=30/36=5/6 ，故選\bbox[red,2pt]{(C)}$$

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解： $$E[Z]=\sum{zp(z)}=\sum{xyf(x,y)} = 2\cdot 1\cdot f(2,1) + 2\cdot 3\cdot f(2,3) + 2\cdot 5\cdot f(2,5) + 4\cdot 1\cdot f(4,1)\\ \;\;+ 4\cdot 3\cdot f(4,3)+ 4\cdot 5\cdot f(4,5) =0.2+ 1.2+1 +0.6+ 3.6+ 3 =9.6，故選\bbox[red,2pt]{(C)}$$

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解： $$E[X]=\sum{xp(x)}=0\times (1/3)+1\times (2/3)= 2/3，故選\bbox[red,2pt]{(C)}$$

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考選部未公布申論題答案，解題僅供參考